Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x+y)=\frac{f(2x)+f(2y)}{2}$ and $f(2x)=2f(x)-1($ put $y=0)$ Now $f^1(x)={\mathrm{Lim}}_{h\to 0}\frac{f(x+h)-f(x)}{h}$

$$\begin{array}{rlr}& ={\mathrm{Lim}}_{h\to 0}\frac{f(2x)+f(2h)-2f(x)}{2h}={\mathrm{Lim}}_{h\to 0}\frac{f(2h)-1}{2h}& =f^1(0)=-1\end{array}$$

/home/mod\_jklog/mod\_jk.log since $f(0)=1$

$$\therefore f(x)=1-x\text{ and }5-f(2)=5-(-1)=6$$